to which we arrive, will have a factor common to all its terms, and consequently will not be the most simple, which may be obtained.

85. Bézout has given a very simple method for exterminating at once all the unknown quantities except one, and for reducing the question immediately to equations, which contain one unknown quantity less, than the equations proposed. Although this process is necessary, only when equations with three unknown quantities are employed, we shall, in order to give a complete view of the subject, begin by applying it to those, which contain only two.

Let there be the equations

ax+by=c

ax + by + c;
y+c;

multiplying the first by any indeterminate quantity m, we have amx+bmy=mc;

subtracting from this result the equation

there remains

a' x + b' y = c',

amx―ax+bmy-by=cm-c',

or (am—a') x + (b m — b') y = cm-c'.

Since m is an indeterminate quantity, we may suppose it to be such, that bm=b'. In this case, the term multiplied by y disappears, and we have

If, instead of supposing bm= b', we make ama', the term, which contains x, will vanish, and we shall have

The value of m will not be the same as before; for we shall have

a'

m

and by substituting this in the expression for y, we find

ca'-ac'

y=ba' — a b'

If we change the signs of the numerator and denominator of this value of y, the denominator will become the same, as that in the expression for æ, since we shall have

a c'. - ca' y=ab-ba a b' — ba'®

86. Next let there be the three equations

a x + by + c z = d

a' x + b'y + c'z = ď

a" x+b" y+c" x=d";

we shall be led, by an obvious analogy, to multiply the first of these equations by m, and the second by n, m and n being indeterminate quantities, to add together the results, and from the sum to subtract the third; by this means, all the equations will be employed at the same time, and the two new quantities m and n, which we may dispose of, as we please, will admit of any determinate value, which may be necessary to make both the unknown quantities to disappear in the result. Having proceeded in this manner, and united the terms by which the same unknown quantity is multiplied, we shall have

(am+a'n — a") x + (b m + b′ n—b′′) y+(cm+c′n—c') ≈

=dm+dnd".

If we would make the unknown quantities x and y to disappear, we must take the equations

am+a'n=a′′

From the two equations, in which m and n are the unknown quantities, it is easy to deduce the value of these quantities, by means of the results obtained in the preceding article; for it is only necessary to change in these results x into m, y into n, and to write instead of the letters

which gives

a, b, c
a', b', c'

}

a, a', a"

the letters {6, U, b",

Substituting these values in the expression for %, and reducing all the terms to the same denominator, we have, (a)

༧=

d (b' a" — a' b'') + d' (a b′′ — b a'') — d" (a b' — b a')

c (b' a'' —a'b'') + c′ (a b' — ba′′) — c" (a b′ — b a')'

If we had made the terms containing x and ≈ to disappear, we should have had y; the letters m and n would have depended upon the equations

am+a'n=a′′ em + cn=c",

and proceeding as before, we should have found

d (c'a" a' c') + d' (a c"-ca")-d" (a c'-ca') y= b (c' a'' — u' c'') + b′ (a c'' — c a') — b′′ (a c' — c a')' Lastly, by assuming the equations

we make the terms multiplied by y and z to disappear; and we have

x=

d (c' b'' — b' c'') + d' (b c" — c b'') — d′′ (b c' — c b')

"

a (c' b'' — b′ c'') + a' (b c" — c b'') — a′′ (b c' — c b′)° These values being developed in such a manner, as to make the terms alternately positive and negative, if we change, at the same time, the signs of the numerator and denominator, in the first and third, we shall give them the following forms;

if we multiply the first by m, the second by n, the third by p, and from the sum of their products subtract the fourth, we shall have

(am + a' n + a" p➡a") x + (b m + b′ n + b′′ p — b′′") y + (cm + c' n + c" p — c") z + (d m+ d'′n+d′′ p—d""') u =em + e'n +e" p—e".

In order to obtain u, we make

The preceding equations, which must give m, n, and p, may be resolved by means of the formulas found for the case of three unknown quantities. This method will appear very simple and convenient; but the nature of the results obtained above will furnish us with a rule for finding them without any calculation.

88. To begin with the most simple case, we take an equation with one unknown quantity, a x = b; from this we find

b x=

a

in which the numerator is the whole known term b, and the denominator the coefficient a of the unknown quantity.

The denominator in this case is composed also of the letters a, a', b, b', by which the unknown quantities are multiplied. We first write a by the side of b, which gives a b; we then change the order of a and b, and obtain b a; prefixing to this the sign we have a b―ba; lastly we place an accent over the last letter in each term, and the expression becomes a b'-ba' for the denominator.

From this expression we may find the numerator. To obtain that for x, we have only to change each a into c, and each b into e for that of y, putting an accent over the last letter as before; in this way we find c b'-bc' for the one, and a c′-ca' for the

other. The numerator may therefore be found from the denominator, as well in cases where there are two unknown quantities, as when there is only one, by changing the coefficient of the unknown quantity sought, into the known term or second member, and retaining the accents, which belonged to the coefficients.

The same rule may be applied to equations with three unknown quantities, as we shall see by merely inspecting the values, which result from these equations. With respect to the denominator, it is necessary further to illustrate the method by which it is formed. Now, since in the case of two unknown quantities, the denominator presents all the possible tranpositions of the letters a and b, by which the unknown quantities are multiplied, it may be supposed, that when there are three unknown quantities, their denominator will contain all the arrangements of the three letters a, b, c. These arrangements may be formed in the following manner.

We first make the transpositions ab-ba with the two letters a and b, then after the first term a b, write the third letter c, which gives a b c; making this letter pass through all the places, observing each time to change the sign, and not to derange the order in which a and b respectively stand, we obtain

abc-acb+cab.

Proceeding in the same manner with respect to the second term -ba, we find

-bac+beacba;

connecting these products with the preceding, and placing over the second letter one accent, and over the third two, we have

ab'c"-adb" +ca'b" —ba' c"+bc' a" —cb' a",

a result, which agrees with that presented by the formulas, obtained above.

From this it is obvious, that, in order to form a denominator in the case of four unknown quantities, it is necessary to introduce the letter d into each of the six products

abc-acb+cab-bac+bca-cba,

and to make it occupy successively all the places. The term abc, for example, will give the four following;

abcd-abde+adbc-dabc.

If we observe the same method in regard to the five other products, the whole result will be twenty four terms, in each of